How do I prove the following: if a+m=b+n;then a=b and m=n(m∈R;n∈R)?

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How do I prove the following: if a+m=b+n;then a=b and m=n(m∈R;n∈R)?

ysnlm8eut · Accepted Answer

When m is squarefree, we know that m is irrational. If we write c=a−b, we getc+m=nand therefore c2+2cm+m=nIf c≠0, we obtainm=n−m−c22cAssuming you mean that a and b are rational numbers, otherwise the claim is clearly false, we deduce a contradiction to m being irrational.Therefore c=0 and, consequently, m=n

ktetik7aeg · Accepted Answer

As posted this is not true - indeed 0+16=2+4. What might be true is that if a,b,m,n∈Z and m,n are real, but not integer, then a=b and m=n.Notice that this is equivalent to m=n implying m=n,where x=x−[x] i.e. the “fractional” part of a number.In any case, let k=a−b and WLOG k≥0. Then k+m=n. If k=0, we are clearly done; suppose k&amp;gt;0. Square both sides to get k2+m+2km=n or equivalently m=n−m−k22k. The last expression is a rational number so m is rational. However, the only way for the square root of an integer to be a rational number is if the square root is an integer, which contradicts our supposition that m is not integer.

How do I prove the following: if

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